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Function Algebra and Geometry

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data

    The function defined on R by f(x)= 2/3x+2

    And cf his graphic representation in an orthonormal mark (o; i; j)

    On the X axis, we consider points A, B and H of respective abscissas 3, 3 and x (x belongs R)

    Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x


    1°) Make the complete figure ?

    2°) Calculate the distances AH and MH according to x and by means of the absolute values ?

    3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?

    2. Relevant equations



    3. The attempt at a solution

    You can make me the graphic representation and the rest I shall make him(it) alone please make for me please
     
  2. jcsd
  3. Feb 28, 2008 #2

    HallsofIvy

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    For clarification, by f(x)= 2/3x+2, do you mean 2/(3x)+ 2 or 2/(3x+2) or (2/3)x+ 2. I suspect the last.

    I am not at all clear what you mean by "cf his graphic representation in an orthonormal mark (o; i; j)". o is the origin, i the x-axis, j the y-axis?
     
  4. Feb 28, 2008 #3
    I mean (2/3)x+2 and o is the origin i the x-axis j the y-axis

    Cf is the name of the curve which represents f(x) = (2/3)x+2

    Can you make this representation please please...

    I ask to you : Can you make this graphic representation ? please just the graphic
     
    Last edited: Feb 28, 2008
  5. Feb 28, 2008 #4

    HallsofIvy

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    Assuming this is f(x)= (2/3)x+ 2, then its graph is a straight line, passing through (0, 2) and (3, 4).

    ?? A and B are on the x-axis and both have abscissa 3? Then A and B are the same point?

    I'm not certain I am interpreting this right but I think, since A= (3, 0), C= (3, 4).
    Since H= (x,0), M= (x, (2/3)x+ 2).

    As I said above, draw the straight line passing through points (0, 2) and (2, 4).

    Okay. A= (2, 0) and H= (x, 0) so AH= |x-2|. M= (x, (2/3)x+ 2). MH= |x- (2/3)x-2|= |(1/3)x- 2|

    The triangle CHA is a right triangle with height AH and Base AC- and AC= |4-2|= 2.
    Area is (1/2)base*height

    The triangle MOB (O is the origin, (0, 0)?) is not a right triangle with but we can still calculate its height MH above, and base OH= x. Again area= (1/2)base*height.

    Set the two areas equal and solve for x.
    2. Relevant equations



    3. The attempt at a solution

    You can make me the graphic representation and the rest I shall make him(it) alone please make for me please[/QUOTE]
     
  6. Feb 28, 2008 #5
    Function

    A has got 3 as abscissa and B has got -3 as abscissa i am sorry i mistake
     
  7. Feb 28, 2008 #6
    Fucntion geometry

    Please can you draw me this curve please please and alla other questions i will do alone thank you help me.
     
  8. Feb 28, 2008 #7
    Function and Geometry (Part 2 )

    The function defined on R by f(x)= 2/3x+2

    And cf his graphic representation in an orthonormal mark (o; i; j)

    On the X axis, we consider points A, B and H of respective abscissas 3, -3 and x (x belongs R)

    Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x


    1°) Make the complete figure ?

    2°) Calculate the distances AH and MH according to x and by means of the absolute values ?

    3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?


    Please Anyone for help me, Can u drwan this graphic please please and represents all points A B H M and C
     
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